# Math Help - Polar Graph question..

1. ## Polar Graph question..

Consider r= cos2theta in polar form.
I understand that if you want the area of one loop you would integrate ( o.5r^2) between -pi/4 to pi/4.

But what is the interpretation of integrating between -pi to pi?
I know it is something to do with r becoming negative but what am i working out in this case?
Also do negative r areas cancel out positive r areas like they do in cartesian?

2. Originally Posted by rodders
Consider r= cos2theta in polar form.
I understand that if you want the area of one loop you would integrate ( o.5r^2) between -pi/4 to pi/4.

But what is the interpretation of integrating between -pi to pi? ... the area of all four petals

Also do negative r areas cancel out positive r areas like they do in cartesian? they can, when you get an "inner loop" ... graph the cardioid r = 0.5 + cos(t) to see what I'm talking about.
...

3. How can areas cancel each other out when you are essentially integrating something squared?
In which case finding the area of the cardoid 0.5 + 1cos(theta)...you would be wrong to integrate between -pi and pi because of the inner loop?

4. ^ Nobody got an ideas here? Its really puzzling me? Ican't see how the area can be negative for r^2 without messing with the limits? But why would you?So how can a inner loop have a negative area?

5. My error ... the inner loop gets counted twice as a positive value, i.e. not negative. The only way I can see to get a negative area is to change the direction of rotation of the sweeping ray , i.e. from $\pi$ to $-\pi$